anonymous
  • anonymous
A floor has two square-shaped designs. The area of the second square-shaped design is nine times greater than the area of the first square-shaped design. Which statement gives the correct relationship between the lengths of the sides of the two squares?
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
these are the options... The length of the side of the second square is 3 times greater than the length of the side of the first square. The length of the side of the second square is 12 times greater than the length of the side of the first square. The length of the side of the second square is 9 times greater than the length of the side of the first square. The length of the side of the second square is 6 times greater than the length of the side of the first square.
anonymous
  • anonymous
since \(A=l^2\) and \(A'=l'^2\) when \(A=9A'\) sub them in to get the relationship between l and l'
anonymous
  • anonymous
what am i supposed to sub in?

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anonymous
  • anonymous
the l and l ' like this: \(A=9A'\) \(l^2 = 9 l'^2\)
anonymous
  • anonymous
im like really slow and have no idea wut im doing.-_-
anonymous
  • anonymous
lol. i guess i'll try to do it step by step. since \(A=l^2\), \(A_1=l_1^2\) (this is the first square)............(1) \(A_2=l_2^2\) (this is the second square)........(2) since the area of the second square is nine times the first, (it's the only given condition, so we start from that.) \(A_2=9A_1\) subbing (1) and (2) into it, we get, \(l_2^2=9l_1^2\)| what did you get after taking thesquare roots of both sides?
anonymous
  • anonymous
sorry im back and thank you.
anonymous
  • anonymous
Just take the square root both the sides and tell us what you got as @Shadowys said above..
anonymous
  • anonymous
Getting ?? @Brianna9898
anonymous
  • anonymous
I have no idea!
anonymous
  • anonymous
Are you getting till here" \[l_2^2=9 l_1^2\]
anonymous
  • anonymous
why is there a 1 & 2 at the bottom
anonymous
  • anonymous
See when you will take square root you will get like: \[\large \sqrt{l_2^2} = \sqrt{9 l_1^2}\] Can you tell what is this: \[\large \sqrt{l_2^2} = ??\]
anonymous
  • anonymous
... and how am I supposed to square root an l ??
anonymous
  • anonymous
Oh that.. \(l_1\) is showing length of first square. \(l_2\) is for length of second square.
anonymous
  • anonymous
you can actually.. See, What is square root of this: \[\large \sqrt{2^2} = ??\]
anonymous
  • anonymous
2
anonymous
  • anonymous
Similar way what will be square root of this: \[\large \sqrt{l_2^2} = ??\]
anonymous
  • anonymous
\[l\]
anonymous
  • anonymous
??
anonymous
  • anonymous
Or you can say \(l_2\).. Good..
anonymous
  • anonymous
Similarly can you tell for: \[\large \sqrt{9l_1^2} = ??\]
anonymous
  • anonymous
so you have to keep the 2 at the bottom
anonymous
  • anonymous
the one and two are sub scripts to differentiate between the first length and the second, thus, 1 and 2 respectively.
anonymous
  • anonymous
see, 1 and 2 is differentiating lengths of the two square you are given with, so don't think here of just l think here of \(l_1\) and \(l_2\)..
anonymous
  • anonymous
so it would be \[9l _{1} ??\]
anonymous
  • anonymous
You forgot to take square root of 9. \(l_1\0 is good though..
anonymous
  • anonymous
What is square root of 9?
anonymous
  • anonymous
3
anonymous
  • anonymous
Yep after square root you will get like: \[\large l_2 = 3 \times l_1\]
anonymous
  • anonymous
So, which answer choice is this?
anonymous
  • anonymous
the first one?
anonymous
  • anonymous
Well Done...
anonymous
  • anonymous
And give all the thanks to @Shadowys
anonymous
  • anonymous
yayyy! thank you guys for helping me!
anonymous
  • anonymous
I don't even know how to give a medal on this thing
anonymous
  • anonymous
Are you seeing best response after shadows post ?/ Just click that..
anonymous
  • anonymous
oh okay, can I give a medal to two ppl?
anonymous
  • anonymous
No, just only one..
anonymous
  • anonymous
On one post you can give medal to one only..
anonymous
  • anonymous
thanx for the medal @Shadowys !

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